Tough Counting

knight247 wrote:How many three digit even numbers N can be formed that are divisible by 9 such that both N and N² have the same units digits?
(A)20
(B)21
(C)22
(D)23
(E)24

OA is A. Detailed explanations would be appreciated.


Another problem can be derived from the above as follows,
How many three digit even integers N can be formed that are divisible by 9 such that both N and N² leave the same remainder when divided by 10?
(A)20
(B)21
(C)22
(D)23
(E)24

Both questions here are asking the exact same thing.

First, for an even number, N, to have the same units digit as N^2, then the units digit of N must be either 6 or 0

Next, there are 900 3-digit integers (from 100 to 999).

Of these 900 integers, every 9th integer is divisible by 9. So, there are 100 3-digit integers that are divisible by 9.

Aside: let's list a few of these integers: 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198, . . .
Notice that, in this list, each units digit is equally represented.

Now, of the 100 integers that are divisible by 9, we are interested in counting only those that end in a 0 or a 6.

Since there are 10 digits altogether (0,1,2,3,4,5,6,7,8,9), then 2/10 of all 100 integers will end in a 0 or a 6.

So the answer here is [spoiler]20 (A)[/spoiler]

Cheers,
Brent